Need help understanding this equation.

Discussion in 'Homework Help' started by zachristensen, Jan 10, 2016.

  1. zachristensen

    Thread Starter New Member

    Jan 10, 2016
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    A 100 mH inductor is placed in parallel with a 100 Ω resistor in a circuit. The circuit has a source voltage of 30 VAC and a frequency of 200 Hz. What is the current through the inductor (rounded)?
     
  2. sailorjoe

    Member

    Jun 4, 2013
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    Zachary, what have you tried so far? Can you draw the circuit? If you can draw it, can you calculate anything about it? Let's start with that.
     
  3. Papabravo

    Expert

    Feb 24, 2006
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    I don't see any equations. Was there a specific one that you were having trouble with?
     
  4. zachristensen

    Thread Starter New Member

    Jan 10, 2016
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    first i found the inductive reactance= xl=2*3.14*200Hz*.1H
    xl=125.6
    next i have to find impedance and thats what im having trouble with
     
    Last edited: Jan 10, 2016
  5. sailorjoe

    Member

    Jun 4, 2013
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    OK good. You have a resistor in parallel with the inductor. And you have the impedance of the inductor at the frequency of the source. Did you know that resistance is like impedance, they are both measured in ohms.
    Do you know how to calculate the equivalent resistance for two resistors in parallel?
     
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  6. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Impedance of inductor is sL=jwL.
    w=2*pi*f=2*pi*(200 Hz)=1256.64 rad/s
    Zl=jwL=j*1256.64*(0.1 H)=125.66j Ohm (this means that your inductor just became a resistor)

    Since Zl, a resistor, is in parallel with a 100 Ohm resistor, you have a current divider. Do you know how to find currents in the current divider?


    A little off topic.
    It has been my experience that reactance is very useless thing. If your instructors require you to find reactance, then go and find it. If they don't ask you to find reactance, then don't waste your time on finding this stupid [impolite term removed].
     
    Last edited by a moderator: Jan 11, 2016
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  7. RBR1317

    Active Member

    Nov 13, 2010
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    Anytime there are components in parallel driven by an ideal voltage source, the current flowing through each component may be calculated independently, as though none of the other components were present.

    If all of the parallel components are the same type (resistor, inductor, or capacitor) then the magnitude of the total current flowing in the circuit is the sum of the individual current magnitudes. However, if the parallel components are of mixed type (which is the case in this problem) then the total current must be calculated as the sum of phasor currents where the phase angle of the individual currents is accounted for. But the question did not ask for the total current in the circuit, so a phasor calculation is not necessary here.
     
    Last edited: Jan 10, 2016
  8. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    This is a current divider yes, but no reason to bring that up because the voltage will drive the current through the inductor without any influence by the resistor because there is no source impedance.
    Therefore the current through the inductor is caused by the voltage only and so can be treated as a reactance: iL=Vs/xL
    and also: iR=Vs/R

    Also, is language like that allowed on this forum? No need for that though even if it is, right?
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    There are some fundamental concepts that, if you understand them, make this problem much simpler.

    The biggest is that you have two components in parallel being driven by a voltage source. That means that they are effectively isolated from each other and have no influence on each other. Hence if you are only interested in finding the current through one of them, you can ignore the other.

    Next it is important to know what is being asked for when the term "current" is used? Are you only being asked to find the magnitude of the current (this is my guess) or are you being asked to find the magnitude and the phase of the current?

    Finally, what math tools do you have in your bag? Are you working with complex impedance, or do your present skills limit you to reactances and vector impedances?
     
  10. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    WBahn brings up good point.
    If OP is asked to find full value of the current, magnitude and phase, then reactance is useless because it only gives you magnitude.

    Frankly, I do math using complex numbers on my calculator, then tell the calculator to produce magnitude or phase or both, as needed, and once again I don't bother with reactance since the complex value contains everything I might need.
     
  11. zachristensen

    Thread Starter New Member

    Jan 10, 2016
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    Here is what I did. unfortunately it does not match the answers that were given. so what did i do wrong?
     
  12. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    It looks like you were trying to calculate the current though both the resistor and inductor.
    The current through the inductor is due to the voltage alone and has nothing to do with the resistor because a voltage source does not change voltage while powering a load unless it has internal resistance. So it is always 30vac.

    The current through the inductor is:
    I=V/sL

    where s is allowed to be changed to j*w, and since it is all imaginary the magnitude is just V/xL.
    Try again and see what you get.

    You probably should also try to visualize how the current flows through these two elements, and why it flows like that. That will help you with future problems too.
     
  13. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    That's actually probably a good habit. That way you are not surprised when there is both real and imaginary parts because you'll be doing it the same way all the time.
     
  14. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    If the point of the exercise to find total current, then:
    Ir=30/100=.3 A
    Il=30/(125.66j)=-0.239j A
    I=Ir+Il=0.3+(-0.239j)=0.3-0.239j A
    Using my calculator:
    Magnitude of I is 0.383 A.
    Phase of I is -38.54°.
     
  15. WBahn

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    Mar 31, 2012
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    As stated in the OP, the question is ONLY asking for the current through the inductor.
     
  16. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Well... Il is 0.239 A @-90°.
     
  17. WBahn

    Moderator

    Mar 31, 2012
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    What were the answers that were given? We are not mind readers.
     
  18. Quartz

    New Member

    Jan 14, 2016
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    Solve the problem in frequency (or phasor) domain.

    Recall:
    100 [Ω] --- in frequency domain ---> 100 [Ω]
    100 [mH] --- in frequency domain ---> j * w * 100 [mH] = j * (2*π*200) * 100 [mH] = j125.66 [Ω]

    So, in frequency domain...
     
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